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givi [52]
3 years ago
15

(a) How can a driver steer a car traveling at constant speed so that the acceleration is zero? (Assume that the road is level. S

elect all that apply.) on a circular path on a straight line on a parabolic path on an elliptical path Incorrect: Your answer is incorrect. (b) How can a driver steer a car traveling at constant speed so that the magnitude of the acceleration remains constant. (Assume that the road is level. Select all that apply.) on a circular path on a straight line on a parabolic path on an elliptical path Incorrect: Your answer is incorrect.
Physics
1 answer:
PilotLPTM [1.2K]3 years ago
8 0

Explanation:

(a) We know that the acceleration of the car is given by :

a = change in speed / time taken

If the speed of the car is constant in a straight line, the acceleration of the car is zero because there is no change in the speed of the car.

(b) For the driver steer a car traveling at constant speed so that the magnitude of the acceleration remains constant, the driver should drive the car in the circular path. This is because, in circular path the speed of an object remains the same while its velocity changes.

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Pand S waves are both
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Answer:

3.body

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The rear window in a car is approximately a rectangle, 1.3 m wide and 0.30 m high. The inside rear-view mirror is 0.50 m from th
maria [59]

Answer:

Height of mirror 0.075 m

width of mirror 0.325 m  

Explanation:

given data

wide = 1.3 m

high = 0.30 m

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solution

we take here height / width of the mirror  is

height / width   = \frac{h}{w}   .................1

and

height /width of the window is

height /width  = \frac{h_w}{w_w}   .................2

and

distance of eye / window by the mirror is

distance of eye / window = \frac{x_e}{x_w}      .................3

so here

θ = θi  = θr    ....................4

and  tanθ for vertical is

tanθ  = \frac{h}{x_e}  

tanθ  =  \frac{h_w}{(x_e + x_w)}       ....................5

so

h =   h_w \times  \frac{x_e}{(x_e + x_e)}     ....................6

put  here value and we get

h = 0.30 \times  \frac{0.50}{(0.50 + 1.50)}  

h = 0.075 m

and

when we take here tanθ for horizontal than it will be

tanθ = \frac{w}{x_e}    

tanθ = \frac{w_w}{(x_e + x_w)}       .......................7

so

w = w_w \times  \frac{x_e}{(x_e + x_w)}         ....................8

put here value and we get

w = 1.3 \times  \frac{0.50}{(0.50 + 1.50)}  

w = 0.325 m

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